What is the extreme f(x)=x^2-2x,1<= x<= 3 ?

The extreme f(x)=x^2-2x,1<= x<= 3 is Minimum(1,-1),Maximum(3,3)

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extreme f(x)=x^2-2x,1<= x<= 3

extreme points

f (x) = x^{2} - 2 x, 1 \leq x \leq 3

Minimum (1, - 1), Maximum (3, 3)

Solution steps

Find the critical points:

x = 1, x = 3

Domain of

x^{2} - 2 x, 1 \leq x \leq 3 : 1 \leq x \leq 3

Find intervals:

Increasing

: 1 < x < 3

Plug the extreme point

x = 1

into

x^{2} - 2 x \Rightarrow y = - 1

Minimum (1, - 1)

Plug the extreme point

x = 3

into

x^{2} - 2 x \Rightarrow y = 3

Maximum (3, 3)

Minimum (1, - 1), Maximum (3, 3)

f (x) = x^{2} - 2 x, 1 \leq x \leq 4

f (x) = \frac{7 x}{x - 4}

21 x^{5} + 35 x^{4}

f (x) = \frac{x + 1}{x ^{2} - 8 x - 9}

f (x, y) = x^{2} + 4 y^{2} + 6

What is the extreme f(x)=x^2-2x,1<= x<= 3 ?
The extreme f(x)=x^2-2x,1<= x<= 3 is Minimum(1,-1),Maximum(3,3)